Question

13.60 Streams of hydrogen \(\left(\mathrm{H}_{2}\right)\) and oxygen \(\left(\mathrm{O}_{2}\right)\), each at \(1 \mathrm{~atm}\), enter a fuel cell operating at steady state and water vapor exits at \(1 \mathrm{~atm}\). If the cell operates isothermally at (a) \(300 \mathrm{~K}\), (b) \(400 \mathrm{~K}\), and (c) \(500 \mathrm{~K}\), determine the maximum theoretical work that can be developed by the cell in each case, in kJ per kmol of hydrogen flowing, and comment. Heat transfer with the surroundings takes place at the cell temperature, and kinetic and potential energy effects can be ignored.

Short answer

The maximum theoretical work is 228,570 kJ/kmol of hydrogen for all temperatures (300 K, 400 K, 500 K).

Step-by-step solution

- Understand the problem and known values

Streams of hydrogen \( \text{H}_{2} \) and oxygen \( \text{O}_{2} \) enter a fuel cell at 1 atm. Water vapor exits at 1 atm. Isothermal operation at temperatures of 300 K, 400 K, and 500 K is given. We are to calculate the maximum theoretical work in each case. We know there is heat transfer at cell temperature, and kinetic and potential energy effects are negligible.

- Write the balanced reaction

The fuel cell reaction is: \[ \text{H}_{2(g)} + \frac{1}{2} \text{O}_{2(g)} \rightarrow \text{H}_2\text{O}_{(\text{vapor})} \]

- Use Gibb's free energy equation to find the maximum theoretical work

The maximum theoretical work is given by the Gibbs free energy change: \[ W_{\text{max}} = \text{-}\bigg[ \text{G}( \text{H}_2\text{O}(\text{vapor}) ) - \text{G}(\text{H}_{2}(\text{g})) - \frac{1}{2} \text{G}(\text{O}_{2}(\text{g})) \bigg] \] \text{Where G denotes Gibbs free energy.}

- Obtain Gibbs free energy values

We use the standard Gibbs free energy of formation values: \[ G_f^\trlo{\text{H}_2} = 0 \text{ kJ/mol} \] \[ G_f^\trlo{\text{O}_2} = 0 \text{ kJ/mol} \] \[ G_f^\trlo{\text{H}_2\text{O}} (\text{vapor}) = -228.57 \text{ kJ/mol} \]

- Calculate Gibbs free energy change at different temperatures

Since Gibbs free energy change doesn't depend on temperature for ideal gases: \[ \text{At 300K: } W_{\text{max}} = - \big[ (-228.57) \text{ kJ/mol} - 0 \text{ kJ/mol} - \frac{1}{2}(0 \text{ kJ/mol}) \big] = 228.57 \text{ kJ/mol} \] \[ \text{At 400K: } W_{\text{max}} = 228.57 \text{ kJ/mol} \] \[ \text{At 500K: } W_{\text{max}} = 228.57 \text{ kJ/mol} \]

- Determine results in kJ per kmol of hydrogen

Since the question asks per kmol of hydrogen and our calculations are mol-based, convert to kmol by multiplying by 1000: \[ W_{\text{max}} = 228.57 \text{ kJ/mol} \times 1000 \text{ mol/kmol} = 228570 \text{ kJ/kmol} \]

- Comment on the results

The maximum theoretical work developed by the cell is the same across all provided temperatures (300 K, 400 K, and 500 K), indicating that temperature does not affect the Gibbs free energy for this reaction under the given conditions.

Key concepts

Gibbs free energy

Gibbs free energy is a fundamental concept in thermodynamics. It represents the maximum amount of work that can be obtained from a thermodynamic process at constant temperature and pressure. For our fuel cell, the Gibbs free energy is critical as it determines the maximum theoretical work the cell can produce.

The Gibbs free energy for a reaction is given by the equation: \[ \text{G}_{\text{reaction}} = \text{G}_{\text{products}} - \text{G}_{\text{reactants}} \]

The output work from a fuel cell is directly related to the change in Gibbs free energy: \[ W_{\text{max}} = -\big(\text{G}_{\text{reaction}}\big) \]
Therefore, calculating the Gibbs free energy change \( \Delta G \)\ for the fuel cell reaction gives us the maximum theoretical work the cell can develop:

Isothermal process

An isothermal process is a thermodynamic process that occurs at a constant temperature. In the context of our fuel cell, this means that while the chemical reactions occur and work is produced, the temperature remains unchanged.

An isothermal process simplifies the calculations significantly because the internal energy change of an ideal gas is zero if the temperature is constant. Hence, all the work done by or on the system can be equated to changes in other forms of energy (like Gibbs free energy).

For our exercise, the fuel cell operates isothermally at temperatures of 300 K, 400 K, and 500 K. Despite this range in temperatures, the Gibbs free energy change and thus the maximum theoretical work remains the same due to the properties of the gases involved.

Maximum theoretical work

The concept of maximum theoretical work is pivotal in understanding fuel cell efficiency. It represents the maximum amount of usable energy that can be derived from a chemical reaction, minus any losses.

In thermodynamics, this can be calculated using the Gibbs free energy change formula. Specifically for our fuel cell reaction: \[ W_{\text{max}} = -\big[G(\text{H}_2\text{O(vapor)}) - G(\text{H}_2(\text{g})) - \frac{1}{2} G(\text{O}_2(\text{g})) \big] \]

Given the reaction \( \text{H}_{2(g)} + \frac{1}{2} \text{O}_{2(g)} \rightarrow \text{H}_2 \text{O}_{(vapor)} \), and with the standard Gibbs free energy values:

• G_f^\trlo{\text{H}_2\text{O(vapor)}} = -228.57 kJ/mol
• G_f^\trlo{\text{H}_2\text{(gas)}} = 0 kJ/mol
• G_f^\trlo{\text{O}_2\text{(gas)}} = 0 kJ/mol

We find that the Gibbs free energy change (and hence the maximum theoretical work) is 228.57 kJ/mol of hydrogen. When scaled to a kmol basis, this value becomes 228,570 kJ/kmol. This implies that at the given conditions, regardless of temperature (300 K, 400 K, or 500 K), the maximum work the fuel cell can theoretically produce remains constant.

Fuel cell reactions

Fuel cells are devices that convert chemical energy directly into electrical energy through electrochemical reactions. In this exercise, the fuel cell involves the reaction of hydrogen (\(H_2\)) and oxygen (\(O_2\)) to produce water vapor (\(H_2O\)):
\[ \text{H}_2\text{(gas)} + \frac{1}{2} \text{O}_2 \text{(gas)} \rightarrow \text{H}_2\text{O} \text{(vapor)} \]

This reaction is exothermic, meaning it releases energy. Fuel cells work by splitting these reactions into two half-reactions, occurring at the anode and cathode. At the anode, hydrogen is oxidized:
\( \text{2H}_2 \rightarrow \text{4H}^+ + \text{4e}^- \)
At the cathode, oxygen reacts with protons and electrons to form water:
\( \text{O}_2 + \text{4H}^+ + \text{4e}^- \rightarrow \text{2H}_2\text{O} \)
Combining these half-reactions, we derive the full reaction, forming water vapor and releasing energy.

A better understanding of these reactions and their Gibbs free energy changes allows us to assess the efficiency and effectiveness of fuel cells in various temperature conditions, as demonstrated in our exercise. Despite operating at different temperatures, the Gibbs free energy and thus the amount of theoretical work that can be extracted remains consistent for the reaction given.